(a) Let M be the midpoint of AB. Since A′ is the centroid of △FAB, the line through A′ and M is the median of △FAB, and therefore passes through B′. Similarly, the line through C′ and M passes through B′, and the line through D′ and M passes through C′. By SAS similarity, we have △A′MB′∼△BMC′, so A′B′=BC′. By the same argument, we can show that B′C′=CE′, C′D′=EF′, and D′E′=FA′.
(b) Let P be the intersection of the line through A′ and D′ and the line through C′ and E′. Since A′ and D′ are the centroids of △FAB and △DEF, respectively, the line through A′ and D′ is parallel to BC and the line through C′ and E′ is parallel to AB. Therefore, P is the midpoint of BC.
[asy] unitsize(1 cm);
pair A, B, C, D, E, F, M, P;
A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (2,2); F = (2,0); M = (2,1); P = (1,2);
draw(A--B--C--D--E--F--cycle); draw(A'--B'--C'--D'--E'--F'--cycle); draw(A'--D',dashed); draw(C'--E',dashed); draw(P--A',dashed); draw(P--C',dashed);
label("A", A, SW); label("B", B, SE); label("C", C, NE); label("D", D, NW); label("E", E, dir(0)); label("F", F, SW); label("A′", A', SW); label("B′", B', SE); label("C′", C', NE); label("D′", D', NW); label("E′", E', dir(0)); label("F′", F', SW); label("M", M, NW); label("P", P, SW); [/asy]
Since P is the midpoint of BC, we have AP=PB=BC/2. Also, A′B′=BC′=BC/2, so △A′PB∼△B′C′. Therefore, the areas of △A′PB and △B′C′ are equal.
Similarly, we can show that the areas of △A′C′E′ and △B′D′F′ are equal. Therefore, △A′C′E′ and △B′D′F′ have equal areas.
